Answer:
the question is not written correctly 16t2 can't be used in the equation
Step-by-step explanation:
do it it correctly you need to combine the like numbers, subtract the 48 from both sides and then divide the number with the t to both sides.
We get the value of t as:
[tex]t=3+\sqrt{6}\ and\ t=3-\sqrt{6}[/tex]
We are asked to solve the given equation for t.
The quadratic equation in terms of the variable t is given by:
[tex]16t^2-96t+48=0[/tex]
On dividing both side of the equation by 16 we get:
[tex]t^2-6t+3=0[/tex]
Now, we know that any quadratic equation of the type:
[tex]at^2+bt+c=0[/tex] has solution as :
[tex]t=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Here,
[tex]a=1,\ b=-6\ and\ c=3[/tex]
i.e.
[tex]t=\dfrac{-(-6)\pm \sqrt{(-6)^2-4\times 1\times 3}}{2\times 1}\\\\t=\dfrac{6\pm \sqrt{36-12}}{2}\\\\t=\dfrac{6\pm \sqrt{24}}{2}\\\\t=\dfrac{6\pm 2\sqrt{6}}{2}\\\\t=3\pm \sqrt{6}[/tex]
i.e.
[tex]t=3+\sqrt{6}\ and\ t=3-\sqrt{6}[/tex]
